SOLUTION: I need to find all zeros 13x^3-185x^2+608x-580

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Question 548737: I need to find all zeros
13x^3-185x^2+608x-580

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I would first look for a rational zero and divide.
A rational zero would be a fraction whose numerator divides 580 and whose denominator divides 13.
The prime factorization of 580 is 580=2%5E2%2A5%2A29
That means it has 12 positive divisors: 1,2,4,5,10,20,29,58,116,145,290, and 580.
Considering the possible denominators 1 and 13 (divisors of 13), that would make 24 positive rational numbers. With the corresponding 24 negative ones, that's 48 possible zeros to try.
Luckily, 2 is a zero of that polynomial. Dividing by x-2
I find 13x%5E3-185x%5E2%2B608x-580=%28x-2%29%2813x%5E2-159x%2B290%29
Applying the quadratic formula to 13x%5E2-159x%2B290=0
we can find any remaining zeros.

That gives us x=10 and x=29%2F13 as the last two zeros.